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I was wondering about important/famous mathematical constants, like $e$, $\pi$, $\gamma$, and obviously the golden ratio $\phi$. The first three ones are really well known, and there are lots of integrals and series whose results are simply those constants. For example:

$$ \pi = 2 e \int\limits_0^{+\infty} \frac{\cos(x)}{x^2+1}\ \text{d}x$$

$$ e = \sum_{k = 0}^{+\infty} \frac{1}{k!}$$

$$ \gamma = -\int\limits_{-\infty}^{+\infty} x\ e^{x - e^{x}}\ \text{d}x$$

Is there an interesting integral* (or some series) whose result is simply $\phi$?

* Interesting integral means that things like

$$\int\limits_0^{+\infty} e^{-\frac{x}{\phi}}\ \text{d}x$$

are not a good answer to my question.

Mike Pierce
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Laplacian
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    You can skim [this page](http://mathworld.wolfram.com/GoldenRatio.html), on WolframAlpha; e.g. Eq (12) and (13). – Clement C. Feb 14 '16 at 03:15
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    Related [question](http://math.stackexchange.com/questions/921293/golden-ratio-from-new-formula-perhaps-from-theory-of-modular-units) introducing an infinite product for GR. And [this question](http://math.stackexchange.com/questions/221584/how-many-infinite-series-representations-of-the-golden-ratio-are-in-existence) – Yuriy S Feb 14 '16 at 03:32
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    Also [this](http://math.stackexchange.com/a/563063/11619). Somewhat famous locally :-) – Jyrki Lahtonen Feb 14 '16 at 09:45
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    In principle, any infinite sum can be expressed as an appropriate contour integral; thus, any of the known infinite sums for $\phi$ can be expressed as contour integrals. – J. M. ain't a mathematician Feb 15 '16 at 14:31
  • Hey guys could we get done proofs of these integrals please? – Faraz Masroor Feb 16 '16 at 12:44
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    An interesting place where the golden ratio comes out is in the calculation of the order of convergence of the secant method https://en.wikipedia.org/wiki/Secant_method – mlainz Feb 17 '16 at 02:23
  • @JyrkiLahtonen thank you for that. good god, that answer was a self contained symphony! – MichaelChirico Feb 18 '16 at 05:01
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    We have the following series representation: $$\phi=\frac{1}{2}+\frac{1331}{250} \sum \limits_{n=0}^{\infty} \frac{(2n+1)!}{5^{3n+1}(n!)^2}.$$ – Antonio Hernandez Maquivar Nov 13 '19 at 18:26
  • If $\varphi=\dfrac{1+\sqrt 5}2$ then $$\int_0^1\frac 1{\sqrt x(1+x)}\,\mathrm dx=\int_{\frac{\varphi^{3/2}-2}{\varphi^{3/2}+2}}^\varphi\frac 1{\sqrt x(1+x)}\,\mathrm dx=\frac \pi{2}$$ This is just $\arctan$ manipulation, but I guess it fits here lol – Mr Pie Jul 19 '21 at 21:05

41 Answers41

129

Potentially interesting:

$$\log\varphi=\int_0^{1/2}\frac{dx}{\sqrt{x^2+1}}$$

Perhaps also worthy of consideration:

$$\arctan \frac{1}{\varphi}=\frac{\int_0^2\frac{1}{1+x^2}\, dx}{\int_0^2 dx}=\frac{\int_{-2}^2\frac{1}{1+x^2}\, dx}{\int_{-2}^2 dx}$$

A development of the first integral:

$$\log\varphi=\frac{1}{2n-1}\int_0^{\frac{F_{2n}+F_{2n-2}}{2}}\frac{dx}{\sqrt{x^2+1}}$$

$$\log\varphi=\frac{1}{2n}\int_1^{\frac{F_{2n+1}+F_{2n-1}}{2}}\frac{dx}{\sqrt{x^2-1}}$$

which stem from the relationship $(x-\varphi^m)(x-\bar\varphi^m)=x^2-(F_{m-1}+F_{m+1})x+(-1)^m$, where $\bar\varphi=\frac{-1}{\varphi}=1-\varphi$ and $F_k$ is the $k$th Fibonacci number. I particularly enjoy:

$$\log\varphi=\frac{1}{3}\int_0^{2}\frac{dx}{\sqrt{x^2+1}}$$ $$\log\varphi=\frac{1}{6}\int_1^{9}\frac{dx}{\sqrt{x^2-1}}$$ $$\log\varphi=\frac{1}{9}\int_0^{38}\frac{dx}{\sqrt{x^2+1}}$$ $$\log\varphi=\frac{1}{12}\int_1^{161}\frac{dx}{\sqrt{x^2-1}}$$

πr8
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  • Wow. Did you come up with this by yourself ? – Saikat Feb 14 '16 at 04:25
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    @user230452 Unfortunately not! Stems from the fact that $\text{arcsinh}{\frac{1}{2}}=\log\varphi$, and this connection comes by noting that $x^2-x-1=0\implies \frac{x-\frac{1}{x}}{2}=\frac{1}{2}$ – πr8 Feb 14 '16 at 04:28
  • This is somehow really really beautiful! Esthetically speaking :D Thank you for that interesting result! – Laplacian Feb 14 '16 at 12:27
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    What about $$\int_0^{1/2}\left(\frac{x}{\sqrt{x^2+1}}+3\right)\,dx$$ –  Feb 14 '16 at 18:02
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    +1 for the understatement, the neat answer and the awesome username. I assume you greet other $\pi r8$s by saying "$Ar^k$" for some $k\geq2$. – David Richerby Feb 14 '16 at 18:23
  • @KimPeek Glad you like it (them)! – πr8 Feb 16 '16 at 13:58
  • @YvesDaoust This is nice, though I am generally partial to cases where the integrand is all together in one piece – πr8 Feb 16 '16 at 13:58
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    @DavidRicherby Indeed - though I'm humbled enough by the reception this first integral seems to have received that I might be well-advised to go by $\varphi$r$8$ from here onwards ^^. – πr8 Feb 16 '16 at 14:00
  • @πr8: would you prefer $\frac{x+3\sqrt{x^2+1}}{\sqrt{x^2+1}}$ ? –  Feb 16 '16 at 14:03
  • @YvesDaoust Somewhat - basically, the less the function looks like a sum/composition of atypical functions, the more attractive I generally find it. If I were looking to develop your suggestion, I'd personally lean towards $\frac{2}{\varphi}=\int_0^2\frac{x}{\sqrt{x^2+1}}\, dx$. Mostly a matter of taste though. – πr8 Feb 16 '16 at 14:08
86

In this answer, it is shown that $$ \int_0^\infty\frac{\sqrt{x}}{x^2+2x+5}\mathrm{d}x=\frac\pi{2\sqrt\phi} $$

robjohn
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    Awesome!! A strict link between $\pi$ and $\phi$, I love those things. Thank you! – Laplacian Feb 14 '16 at 14:35
  • Brilliant!! Absolutely amazing – Sayan Chattopadhyay Feb 14 '16 at 15:04
  • wow! this is incredible – Andres Mejia Feb 14 '16 at 16:45
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    So we know $\pi=2e\int_0^{\infty}{\cos(x)\over x^2+1}\text{d}x$ and $e=\sum_{k=0}^{\infty}{1\over k!}$ from the OP, then this answer says $\int_0^\infty{\sqrt{x}\over x^2+2x+5}\text{d}x={\pi\over 2\sqrt{\Phi}}$. My immediate thought was to combine the above to get $\Phi=\left({\sum_{k=0}^{\infty}{1\over k!}\int_0^{\infty}{\cos(x)\over x^2+1}\text{d}x \over \int_0^\infty{\sqrt{x}\over x^2+2x+5}\text{d}x}\right)^2$, which might be considered "interesting". – MichaelS Feb 14 '16 at 22:36
  • So very nice ! Somehow you perhaps can rope in $e$ too. – Narasimham Feb 15 '16 at 15:18
57

An identity derived from the Rogers-Ramanujan continued fraction ($R(q)$, not defined here) exhibits a $\phi$ factor:

$$ \frac{1}{(\sqrt{\phi\sqrt{5}})e^{2\pi/5}} = 1+\frac{e^{-2\pi}}{1+\frac{e^{-4\pi}}{1+\frac{e^{-6\pi}}{1+\frac{e^{-8\pi}}{1+\frac{e^{-10\pi}}{1+\frac{e^{-12\pi}}{\cdots}}}}}}$$

artist view of the identity

and one can then obtain a formula like: $$ \ln \left( \sqrt{4\phi+3}-\phi^2\right) = -\frac{1}{5}\int_{e^{-2\pi}}^1 \frac{(1-t)^5(1-t^2)^5(1-t^3)^5 \dots}{(1-t^5)(1-t^{10})(1-t^{15}) \dots}\frac{dt}{t}$$ which beautifully links integrals, $e$, $\phi$ and $\pi$. It is described for instance in Golden Ratio and a Ramanujan-Type Integral. Not very practical though to obtain $\phi$ rational approximations.

In M. D. Hirschhorn, A connection between $\pi$ and $\phi$, Fibonacci Quarterly, 2015, another asymptotic relation is:

$$ \frac{1}{\pi}=\lim_{n\to \infty} 2n {5}^{1/4}\sum_{k=0}^{n}\binom{n}{k}^2\binom{n+k}{k}/\phi^{5n+5/2}$$

Laurent Duval
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    The genius of Ramanujan will always remain a mystery.. what a genius. – Laplacian Feb 15 '16 at 14:33
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    And I believe it is a good thing that this remains a mystery. – Laurent Duval Feb 15 '16 at 14:58
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    mind... blown... – MichaelChirico Feb 18 '16 at 05:03
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    @FourierTransform right! – Fawad Oct 08 '16 at 09:14
  • The link seems dead, here is an archived version: [pdf](https://web.archive.org/web/20150922074159if_/https://www.mdpi.com/2075-1680/2/1/58/pdf), [html](https://web.archive.org/web/20180131025715if_/https://www.mdpi.com/2075-1680/2/1/58/htm). – Vladimir Reshetnikov Jan 31 '18 at 03:08
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    @VladimirReshetnikov I even made a question on this ([here](https://math.stackexchange.com/questions/2655006/the-most-complex-formula-for-the-golden-ratio-varphi-that-i-have-ever-seen-h)) :D – Mr Pie Sep 01 '18 at 18:39
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    Btw, here is another formula owing to Ramanujan: $$1-\frac {(1!)^2}{2!\times 3^2}+\frac{(2!)^2}{4!\times 5^2}-\frac{(3!)^2}{6!\times 7^2}+\frac{(4!)^2}{8!\times 9^2}-\cdots$$ $$=\frac{\pi^2}6-3\ln^2\varphi$$ – Mr Pie Jul 19 '21 at 21:21
47

For $k>0$, we have

$$\bbox[8pt,border:3px #FF69B4 solid]{\color{red}{\large \int_0^\infty \ln \left( \frac{x^2-2kx+k^2}{x^2+2kx\cos \sqrt{\pi^2-\phi}+k^2}\right) \;\frac{\mathrm dx}{x}=\phi}}$$ I hope you find this integral interesting.

Extra: $$\bbox[8pt,border:3px #FF69B4 solid]{\color{red}{\large \int_0^\infty \frac{x^{\frac\pi5-1}}{1+x^{2\pi}} \mathrm dx=\phi}}$$

Venus
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$$\int_{-1}^1 dx \frac1x \sqrt{\frac{1+x}{1-x}} \log{\left (\frac{2 x^2+2 x+1}{2 x^2-2 x+1}\right )} = 4 \pi \operatorname{arccot}{\sqrt{\phi}}$$

Ron Gordon
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35

Here's a series:

$$ \phi = 1 + \sum_{n=2}^\infty \frac{(-1)^{n}}{F_nF_{n-1}} $$

where $F_n$ is the $n$th Fibonacci number.

To see this, rewrite the numerator using the identity $(-1)^n=F_{n+1}F_{n-1}-F_n^2$, at which point the summand becomes $$ \frac{F_{n+1}F_{n-1}-F_n^2}{F_nF_{n-1}}=\frac{F_{n+1}}{F_n}-\frac{F_n}{F_{n-1}} $$ and so the sum telescopes: the partial sum ending at $n$ is equal to $$ \frac{F_{n+1}}{F_n}-\frac{F_2}{F_1}=\frac{F_{n+1}}{F_n} - 1 $$ which gives the original expression for the series via the limit $\lim_{n \to \infty} \frac{F_{n+1}}{F_n} = \phi$.

Micah
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  • Was this the first definition of golden ratio or did it have a definition before that ? – Saikat Feb 14 '16 at 04:27
  • @user230452 $\phi = \frac { 1+ \sqrt 5}2$ – Ant Feb 14 '16 at 09:43
  • I mean, didn't that number come from the Fibonacci series itself or did it already have a definition and was found again in the Fibonacci series ? – Saikat Feb 14 '16 at 10:23
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    @user230452 The golden ratio first arose as a ratio between quantities $a,b$ for which ratio $a:b$ is the same as ratio $a+b:a$. Hence, it came before (or at least independently of) Fibonacci numbers. – Wojowu Feb 14 '16 at 10:32
  • @Wojowu I agree that the original definition is independent of the Fibonacci sequence, though we're really using "independent" in the sense of copyright law rather than anything else. Fundamentally, it's essentially the same definition. – David Richerby Feb 14 '16 at 18:26
  • @DavidRicherby Well, all definitions of golden ratio are essentially the same definition by that line of thought, because they define the same number. – Wojowu Feb 14 '16 at 21:10
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    @Wojowu My point was just that the Fibonacci sequence is all about situations where you're dealing with $a$, $b$ and $a+b$, and so is the classical definition of the golden ratio. Whereas, for example, $(1+\sqrt{5})/2$ is a completely different way of defining the same number. Anyway, I'm just nit-picking. – David Richerby Feb 14 '16 at 21:26
  • @user230452 the Greeks were very keen on the Golden Ratio. I had always assumed from his name that Fibonacci was much later. https://en.wikipedia.org/wiki/Golden_ratio confirms this. – Level River St Feb 14 '16 at 21:44
31

Based on the fact that $\varphi = \frac{1+\sqrt{5}}{2}$:

$$\varphi = \int_4^5 \frac32+\frac1{4\sqrt{x}} \mathrm{d}x$$

Based on the fact that $\varphi = 2\cos(\frac{\pi}{5})$:

$$\varphi = \int_{\tfrac{\pi}{5}}^{\tfrac{\pi}{2}} 2\sin(x) \mathrm{d}x$$

wythagoras
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$$\int_0^{\infty} \frac{x^2}{1+x^{10}} \, \mathrm{d}x = \frac{\pi}{5 \phi}.$$

user314474
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$$\int_0^{\infty} \frac{dx}{(1+x^\phi)^\phi}=1$$

Vladimir Reshetnikov
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karvens
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22

All the following is based on the simple fact that:

$$\phi=2 \cos \left( \frac{\pi}{5} \right)=2 \sin \left( \frac{3\pi}{10} \right)$$

These integrals are the small sample of what we can build using this identity:

$$\frac{1}{2 \pi} \int_0^{\infty} \frac{dx}{(1+x)x^{0.7}}=\phi-1$$

$$\frac{1}{1.4 \pi} \int_0^{\infty} \frac{dx}{(1+x)^2x^{0.7}}=\phi-1$$

$$\frac{1}{2 \pi} \int_0^{1} \frac{dx}{(1-x)^{0.3}x^{0.7} }=\phi-1$$

$$\frac{5}{3 \pi} \int_0^{1} \frac{x^{0.3}dx}{(1-x)^{0.3} }=\phi-1$$

$$\frac{1}{2 \pi} \int_1^{\infty} \frac{dx}{(x-1)^{0.3}x }=\phi-1$$

$$\frac{1}{0.21 \pi} \int_0^{\infty} \frac{x^{0.3}dx}{(1+x)^{3} }=\phi-1$$

Take any tables of definite integrals, find any one that ends in a trig function and set the parameters to obtain $\phi$.


You can find the following infinite product for $\phi$ here

$$2 \phi=\prod_{k=0}^{\infty}\frac{100k(k+1)+5^2}{100k(k+1)+3^2}$$

It's converging slowly, see the link for the proof using the properties of Gamma function.

By numerical computation at $50000$ terms this infinite product gives only $5$ correct digits for $\phi$, giving $1.618029$ instead of $1.618034$.

Using the infinite product for $\cos(x)$, we get:

$$\frac{\phi}{2}=\prod_{k=1}^{\infty}\left(1- \frac{4}{5^2 (2k-1)^2} \right)$$

This infinite product at $50000$ terms gives $\phi=1.618035$, only $4$ correct digits. This is actually almost the same product, because if we rearrange it we get:

$$\frac{\phi}{2}=\prod_{k=0}^{\infty}\left(\frac{100 k (k+1)+21}{100 k (k+1)+25} \right)$$

I suggest looking at this question for much more interesting product.

Yuriy S
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The length of the logarithmic spiral $\rho=e^{2\theta}$ up to $\theta=0$ is given by

$$\int_{-\infty}^0\sqrt{\rho^2+\dot\rho^2}d\theta=\int_{-\infty}^0\sqrt{1+2^2}e^{2\theta}d\theta=\phi-\frac12.$$

17

How about this one:

$$\int_0^1 \frac{dx}{\sqrt{x+\sqrt{x+\sqrt{x+\cdots}}}}=\frac{2}{\phi}-\ln \phi$$

There is an infinitely nested radical in the denominator.

A finite one is also possible:

$$\int_0^{1/16} \frac{dx}{\sqrt{x+\sqrt{x}}}=\phi-2\ln (\phi)-\frac12$$

Yuriy S
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I am not taking credit for this. I am just posting this because it answers the question. I give Felix Marin and Olivier Oloa complete credit for these results.

$$\int_0^{\pi/2} \ln(1+4\sin^2 x)\text{ d}x=\pi\log\left(\varphi\right)$$

and

$$\int_0^{\pi/2} \ln(1+4\sin^4 x)\text{ d}x=\pi\log \frac{\varphi+\sqrt{\varphi}}{2}$$

Again, not mine. But they definitely deserve to be here

10

$$\int_0^\infty x(2x-1)\,\delta(x^2-x-1)\,dx$$


Update:

As pointed by Yuriy, we must take into account the derivative of the argument of the $\delta$ function. This is why the corrective factor $2x-1$ appears.

More generally,

$$\int_I x|g'(x)|\delta(g(x))\,dx$$ evaluates to the root of $g$ contained in the interval $I$, provided there is only one. The first factor $x$ can be replaced by any function $f(x)$ to yield the value of that function at the root.

  • Beautiful!! Dirac Delta. Very easy and elegant, thank you! – Laplacian Feb 14 '16 at 17:43
  • A great idea, actually! We can do it for any algebraic number, it seems – Yuriy S Feb 14 '16 at 19:24
  • @YuriyS: yep, provided you isolate the desired root in an interval. –  Feb 14 '16 at 19:36
  • Actually, Wolframalpha gives another value for this integral: http://www.wolframalpha.com/input/?i=integrate+x+DiracDelta%5Bx%5E2-x-1%5D+from+x%3D0+to+infinity – Yuriy S Feb 14 '16 at 20:19
  • In general $\delta [g(x)]=\sum_k \frac{\delta (x-x_k)}{| g'(x_k)|}$ – Yuriy S Feb 14 '16 at 20:25
  • @YuriyS: quite right, I forgot that. Fortunately, we can fix by multiplying by $|g'(x)|$. –  Feb 14 '16 at 20:48
  • @YuriyS: the method extends to all transcendental numbers that can be expressed as a root of a function, such as $\pi$ or $e$ or $W(2)$... –  Feb 14 '16 at 21:14
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$$ \int_0^\infty \frac{1}{1+x^{10}}dx=\frac{\phi\pi}{5} $$

9

Five:Pi:Phi

Coincidence closed form

$$\int_{0}^{\pi\over 2}\mathrm dx \sqrt[5]{\tan(x)}\cdot{\ln(\csc^2(x))\over \sin(2x)}=\color{blue}5\color{red}\pi\color{brown}\phi$$

7

$$ \int_0^1 \frac{1+x^8}{1+x^{10}}dx=\frac{\pi}{\phi^5-8} $$

7

Here is another one

$$\int_{-\infty}^{+\infty}e^{-x^2}\cos (2x^2)\mathrm dx=\sqrt{\phi \pi\over 5}$$

gymbvghjkgkjkhgfkl
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An integral uniting some favourite mathematical constants

$$\int_{-\infty}^{+\infty}\frac{t^2}{(\phi^n t)^2+(F_{2n+1}-\phi F_{2n})(\pi t^2+\zeta(3)t-e^{\gamma})^2}\mathrm dt=1$$

Where,

$\phi$; Golden ratio

$\zeta(3)$; Apery's constant

$\gamma$; Euler-Mascheroni's constant

$e$; Euler Number

$F_{n}$; Fibonacci number

and $\pi=3.14...$

Laplacian
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Consider the sequence

$1,2,2,3,3,4,4,4,...$

where $a_1=1,a_{n+1}\in\{a_n,a_n+1\}$, and $a_n$ is the number of times $n$ occurs in the sequence. Then if we assume that $a_n$ grows asymptotically as $\alpha n^\beta$, we get

$\alpha=\phi^{1/{\phi^2}}$

$\beta=1/\phi$.

I saw this is a textbook problem on asymptotic analysis. It turns out that for all $n$ the asymptotic expression is well within one unit of the actual $a_n$.

Oscar Lanzi
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Here is a collection of the series with reciprocal binomial coefficients.

$$\sum_{n=0}^\infty (-1)^n \left( \begin{matrix} 2n \\ n \end{matrix} \right)^{-1}=\frac{4}{5} \left(1-\frac{\sqrt{5}}{5} \ln \phi \right)$$

$$\sum_{n=1}^\infty \frac{(-1)^n}{n} \left( \begin{matrix} 2n \\ n \end{matrix} \right)^{-1}=-\frac{2\sqrt{5}}{5} \ln \phi$$

$$\sum_{n=1}^\infty \frac{(-1)^n}{n^2} \left( \begin{matrix} 2n \\ n \end{matrix} \right)^{-1}=-2 \ln^2 \phi$$

$$\sum_{n=0}^\infty \frac{(-1)^n}{2n+1} \left( \begin{matrix} 2n \\ n \end{matrix} \right)^{-1}=\frac{4\sqrt{5}}{5} \ln \phi$$

$$\sum_{n=0}^\infty \frac{(-1)^n}{n+1} \left( \begin{matrix} 2n \\ n \end{matrix} \right)^{-1}=\frac{8\sqrt{5}}{5} \ln \phi-4 \ln^2 \phi$$

$$\sum_{n=2}^\infty \frac{(-1)^n}{n-1} \left( \begin{matrix} 2n \\ n \end{matrix} \right)^{-1}=\frac{3\sqrt{5}}{5} \ln \phi-\frac{1}{2}$$

$$\sum_{n=2}^\infty \frac{(-1)^n}{(n-1)^2} \left( \begin{matrix} 2n \\ n \end{matrix} \right)^{-1}=1-\sqrt{5} \ln \phi+ \ln^2 \phi$$

$$\sum_{n=2}^\infty \frac{(-1)^n}{n^2(n^2-1)} \left( \begin{matrix} 2n \\ n \end{matrix} \right)^{-1}=4\ln^2 \phi-\frac{\sqrt{5}}{2} \ln \phi-\frac{3}{8}$$

A one with $\pi$:

$$\sum_{n=0}^\infty \left( \begin{matrix} 4n \\ 2n \end{matrix} \right)^{-1}=\frac{16}{15}+\frac{\sqrt{3}}{27} \pi-\frac{2\sqrt{5}}{25} \ln \phi $$

Source here

Yuriy S
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So you said that series are OK, so I will offer a few:

$$\phi=\frac{13}{8}+\sum_{n=0}^\infty \frac{(-1)^{n+1}(2n+1)!}{n!(n+1)!4^{(2n+3)}}$$

$$\phi=2\cos (\pi/5)=2\sum_{k=0}^\infty \frac{((-1)^k (\pi/5)^{2 k}}{(2k)!}$$

$$\phi=\frac{1}{2}+\frac{\sqrt{5}}{2}=\frac{1}{2}+\sum_{n=0}^\infty 4^{-n}\binom{1/2}{n}$$

5

Not exactly a series, but might also be of interest:

$$1-\frac{1}{\phi}=\frac{1}{\phi^2}=\frac{1}{5} \left(1+\frac{1}{5} \left(1+\frac{1}{5} \left(1+\frac{1}{5} \left(1+\dots \right)^2 \right)^2 \right)^2 \right)^2$$


$$\frac{1}{\phi^4}=\frac{1}{5} \left(1-\frac{1}{5} \left(1-\frac{1}{5} \left(1-\frac{1}{5} \left(1-\dots \right)^2 \right)^2 \right)^2 \right)^2$$

$$\frac{1}{\phi^4}=\frac{1}{9} \left(1+\frac{1}{9} \left(1+\frac{1}{9} \left(1+\frac{1}{9} \left(1+\dots \right)^2 \right)^2 \right)^2 \right)^2$$

Yuriy S
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Let $ F_0=0, F_1=1 ; F_{n+1}=F_{n-1}+F_n $ be the Fibonacci numbers

$\zeta(s)$ is the zeta function. Then:

$$ \prod_{n=1}^{\infty}\left[(-1)^{n+1}\phi F_n+(-1)^nF_{n+1}\right]^{n^{-(s+1)}}=\phi^{-\zeta(s)} $$

wythagoras
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We can prove the inequalities

$$\frac{3}{2}<\frac{8}{5}<\phi<\frac{13}{8}<\frac{5}{3}$$

with representations

$$\begin{align}\phi&=\frac{3}{2}+\frac{1}{4}\int_0^1 \frac{dx}{\sqrt{4+x}}\\ \\ \phi&=\frac{8}{5}+\frac{1}{5}\int_0^1 \frac{dx}{\sqrt{121+4x}}\\ \\ \phi&=\frac{13}{8}-\frac{1}{16}\int_0^1 \frac{dx}{\sqrt{80+x}}\\ \\ \phi&=\frac{5}{3}-\frac{1}{3}\int_0^1 \frac{dx}{\sqrt{45+4x}}\\ \end{align}$$

Jaume Oliver Lafont
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$$\int_0^1\frac{\ln(1+x-x^2)}{1-x}dx=\int_0^1\frac{\ln(1+x-x^2)}xdx=2\ln^2\varphi$$ $$\int_0^1\frac{\ln(1-3x+x^2)\ln x}{x}dx=\frac85 \zeta (3)+\frac{2}{5} \pi ^2 \ln \varphi-2 i \pi \ln^2\varphi$$

Kemono Chen
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$$\int_{-\infty}^{+\infty}\frac{\mathrm dx}{(1+x+x^2)^2+x^2}=\pi\cdot \sqrt{\frac{\phi}{5}}$$

3

-I remember really liking this one:

$$\int_0^1 \int_0^1 \frac{\text{dx dy}}{\varphi^6-x^2y^2}=\frac{\pi^2-18\log^2\varphi}{24\varphi^3}$$

I most liked it because it was specific to $\varphi$

-Also, we can note this M.SE result (with some interpolation)

$$\int_0^1 \frac{\log (1+x^{\alpha+\sqrt{\alpha^2-1}})}{1+x}\text{dx}=$$$$\frac{\pi^2}{12}\left(\frac{\alpha}{2}+\sqrt{\alpha^2-1}\right)+\log(\varphi)\log(2)\log(\sqrt{\alpha+1}+\sqrt{\alpha-1})\log(\text{something})$$

Perhaps someone can help me fill in $\text{"something"}$

3

Here is another one $$ \int_0^\infty \frac{1}{5^{\frac{x}{4}}+5^{\frac{1}{2}}-5^0}dx=\phi $$

3

This one is a bit messy.

$$ \int_0^\infty \frac{1}{(\sqrt5^x)^{2^{-(\sqrt5-1)}}+\sqrt5-1}dx=2^{\phi^{-3}}\cdot\phi $$

3

$$\int_0^\infty \frac{1}{1+x^{\frac{10}{3}}}dx=\frac{3\pi}{5\phi}$$

3

Notice that $\frac{2}{1+\sqrt5}=\frac{1}{\phi}$

$$\int_0^1\frac{2}{(1+\sqrt5x)^2}dx=\frac{1}{\phi}$$

3

$$\int_0^1 \frac{200\sqrt5(1-x^2)-300(1-x)^2}{ \left[5\sqrt5(1+x)^2-15(1-x^2)+2\sqrt5(1-x)^2 \right]^2}dx=(2\phi+1)(\phi+2)$$

3

$$\int_{0}^{\phi}(1-x+x^2)^{1/\phi}(1-\phi^2x+\phi^3x^2)\mathrm dx=2^{\phi}\cdot\phi$$

A bit over-crowed in term of $\phi$

gymbvghjkgkjkhgfkl
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3

Double integral of $\phi$ $$-\frac{1}{5}\int_{0}^{1}\int_{0}^{1}\frac{\mathrm dx \mathrm dy}{\left(\frac{1}{5}-x+x^2\right)\sqrt{1-y+\frac{y^2}{5}}}=\ln\left(\frac{1}{\phi^4}\right)\ln\left(\frac{\phi^2+1}{\phi^4}\right)$$

Without the natural logarithm $$\int_{0}^{1}\int_{0}^{1}\frac{x^3}{(2x^2-2x+1)^2\left(x-\frac{1}{2}\right)^2\sqrt{(\phi+y^2)^3}}\mathrm dx \mathrm dy=-\frac{1}{\phi^2}$$

Sibawayh
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3

By Euler's reflection formula, it follows that

$$ \int_0^\infty{x^{s-1}\over1+x}\mathrm dx={\pi s\over\sin(\pi s)}\tag1 $$

Accordingly, we can find an $s$ such that $\sin(\pi s)$ can be associated with $\phi$. As it turns out, we do have some special angle that allows us to do so.

Image

By observing the geometric properties of this triangle, we can deduce the following relationship

$$ \triangle ABC\sim\triangle BDA\cong\triangle DBA $$

which implies

$$ {BC\over AB}={AB\over BD} $$

Now, due to the properties of isosceles triangles, we get

$$ AB=AD=CD\Rightarrow BC=BD+CD=BD+AB $$

Thus, we obtain

$$ 1+{BD\over AB}={AB\over BD} $$

To convenience the derivation, set $AB=1,BD=y$ so that the above identity becomes

$$ 1+y=\frac1y\Rightarrow y^2+y-1=0\Rightarrow y={-1+\sqrt5\over2}=\frac1\phi $$

Again, by the properties of isosceles, we deduce

$$ CE={1+y\over2} $$

As a result, we obtain $\cos36^\circ$ from its definition:

$$ \cos36^\circ={CE\over CD}={CE\over AB}={1+y\over2}={1+\sqrt5\over4}=\frac\phi2 $$

Now, due to the conversion that

$$ 90^\circ-36^\circ=54^\circ={3\pi\over10} $$

we obtain

$$ \sin\left(3\pi\over10\right)=\frac\phi2 $$

Therefore, setting $s=3/10$ in (1), we obtain

$$ \fbox{$\Large\int_0^\infty{x^{-7/10}\over1+x}\mathrm dx={2\pi\over\phi}$} $$

TravorLZH
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2

$$\int_0^1 \frac{2}{\left(1+\phi x^2\right) \sqrt{1-x^2}} \, dx=\frac{\pi }{\phi }$$

Mariusz Iwaniuk
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1

$$\int_{0}^{1}{1-x^{\phi}\over 1-x}+{x(1-\phi x^{1\over \phi}+{1\over \phi}x^{\phi})\over (1-x)^2}\mathrm dx=\phi$$

gymbvghjkgkjkhgfkl
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1

Here is another one:

$$\int_{0}^{1}(2x-1)\left(\frac{1}{x(1-x)}+\frac{\phi\pi}{(1+x)(2-x)}+\frac{\phi \pi^2}{(2+x)(3-x)}+\frac{\phi \pi^3}{(2+x)(3-x)}+\cdots\right)-\frac{1}{\ln(1-x)}+\frac{x^{e-1}}{\ln x} \mathrm dx=1$$

Sibawayh
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0

enter image description here

This is my attempt to find another integral representation for Golden ratio using some special function as shown in the above image !!!!!!!

zeraoulia rafik
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    I already saw it in a past question of yours. It's really cool, can you prove it? – Laplacian Mar 22 '18 at 06:42
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    Can we see all the proof please ? – Abr001am May 08 '18 at 15:28
  • Does the constant $t$ have any significance other than this? Otherwise, not all that interesting (sorry), since you're just defining $t$ as $t=I^{-1}(\phi)$ and then concluding $I(t) = \phi$. – Dark Malthorp Apr 11 '20 at 15:32
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    e.g. there also exists a constant at $x\approx 2.7573$ such that $\Gamma(x) = \phi$, but this isn't interesting since $\Gamma$ takes on all values in $\mathbb{R}\setminus\{0\}$. – Dark Malthorp Apr 11 '20 at 15:34
0

Connecting three well-known constants together: $$\int_{-\infty}^{+\infty}\sin^2\left(\frac{x}{\phi+\frac{1}{\phi}}\right)\cdot \frac{\mathrm dx}{4x^4+5x^2}=\frac{\pi}{e}\cdot\frac{1}{\left(\phi+\frac{1}{\phi}\right)^3}$$

Sibawayh
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